Let { X i } ∼ N ( i θ , 1 ) for i = 1 , . . . . , n be an independent, but not identically distributed sample. Check that T = ∑ i X i it is a sufficient statistic for θ .

Question

Let   {  X  i  }  ∼  N  (  i      θ    ,  1  ) for   i  =  1  ,  .  .  .  .  ,  n be an independent, but not identically distributed sample. Check that   T  =      ∑    i        X    i   it is a sufficient statistic for       θ  .

Ryan Davies · Accepted Answer

We have that the probability density function of       X    i   is      f                  X        i              (      x    i    )  =            1              2        π              exp  ⁡      [    −                  1        2                            (                  x          i                −        i        θ        )            2        ]    .By independence, the joint density is given by      f                  X        1            ,      …      ,              X        n              (      x    1    ,  …  ,      x    n    )  =      ∏          i      =      1              n            f                  X        i              (      x    i    )  =            1              (        2        π                  )                      n                          /                        2                                exp  ⁡      [    −                  1        2                    ∑              i        =        1                    n              (          x      i        −    i    θ          )      2        ]    .Now we expand the sum:      ∑          i      =      1              n        (      x    i    −  i  θ      )    2    =      ∑          i      =      1              n            x    i    2    −  2  θ      ∑          i      =      1              n        i      x    i    +  θ      ∑          i      =      1              n            i    2    .Recalling the sum      ∑          i      =      1              n            i    2    =                    n        (        n        +        1        )        (        2        n        +        1        )            6      we obtain      ∑          i      =      1              n        (      x    i    −  i  θ      )    2    =      ∑          i      =      1              n            x    i    2    +  θ      [                            n          (          n          +          1          )          (          2          n          +          1          )                6              −    2          ∑              i        =        1                    n              i          x      i        ]    .The joint density may thus be written as                                                        1                              (                2                π                                  )                                      n                                          /                                        2                                                                                exp          ⁡                      [            −                                          1                2                                                    ∑                              i                =                1                                            n                                                    x              i              2                        ]                          ⏟                    h      (              x            )                                    exp          ⁡                      {            θ                          [                                                                    n                    (                    n                    +                    1                    )                    (                    2                    n                    +                    1                    )                                    6                                            −              2                              ∑                                  i                  =                  1                                                  n                                            i                              x                i                            ]                        }                          ⏟                    φ      (              ∑                  i          =          1                          n                    i              x        i            ,      θ      )        .By the factorization criterion,       ∑          i      =      1              n        i      X    i   is sufficient for   θ.      ∑          i      =      1              n            X    i    is sufficient for   θ.

Let (Xi) sim N(i theta, 1) for i=1,....,n be an independent, but not identically distributed sample. Check that T = sum_i X_i it is a sufficient statistic for theta.

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